\newproblem{lay:7_3_1}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 7.3.1}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
	Find the change of variable $\mathbf{x}=P\mathbf{y}$ that transforms the quadratic form $\mathbf{x}^TA\mathbf{x}=\mathbf{y}^TD\mathbf{y}$ as shown
	\begin{center}
		$5x_1^2+6x_2^2+7x_3^2+4x_1x_2-4x_2x_3=9y_1^2+6y_2^2+3y_3^2$
	\end{center}
}{
   % Solution
	Let $A$ be
	\begin{center}
		$A=\begin{pmatrix}5 & 2 & 0 \\ 2 & 6 & -2 \\ 0 & -2 & 7\end{pmatrix}$
	\end{center}
	We may orthogonally diagonalize it as 
	\begin{center}
		$A=PDP^T=\begin{pmatrix}-\frac{1}{3} & \frac{2}{3} & -\frac{2}{3} \\ -\frac{2}{3} & \frac{1}{3} & \frac{2}{3} \\ \frac{2}{3} & \frac{2}{3} & \frac{1}{3}\end{pmatrix}
		  \begin{pmatrix}9 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 3\end{pmatrix}
			\begin{pmatrix}-\frac{1}{3} & \frac{2}{3} & -\frac{2}{3} \\ -\frac{2}{3} & \frac{1}{3} & \frac{2}{3} \\ \frac{2}{3} & \frac{2}{3} & \frac{1}{3}\end{pmatrix}^T$
	\end{center}
	The required change of variable is $\mathbf{x}=P\mathbf{y}$ with
	\begin{center}
		$P=\begin{pmatrix}-\frac{1}{3} & \frac{2}{3} & -\frac{2}{3} \\ -\frac{2}{3} & \frac{1}{3} & \frac{2}{3} \\ \frac{2}{3} & \frac{2}{3} & \frac{1}{3}\end{pmatrix}$
	\end{center}
}
\useproblem{lay:7_3_1}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}

